AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 19 Hard

The diameter AB of a circle of radius 2 is extended to a point D outside the circle so that BD=3. Point E is chosen so that ED=5 and line ED is perpendicular to line AD. Segment AE intersects the circle at a point C between A and E. What is the area of \triangle ABC?

  • A.

    \frac{120}{37}

  • B.

    \frac{140}{39}

  • C.

    \frac{145}{39}

  • D.

    \frac{140}{37}

  • E.

    \frac{120}{31}

Answer:D

We note that \triangle ACB\sim\triangle ADE by AA similarity. Also, since the area of \triangle ADE=\dfrac{7\cdot5}{2}=\dfrac{35}{2} and AE=\sqrt {74},\dfrac{A_{\triangle ABC}}{A_{\triangle ADE}}=\dfrac{A_{\triangle ABC}}{\dfrac{35}{2}}=\left( \dfrac{4}{\sqrt{74}}\right)^{2} , so the area of \triangle ABC=\frac {140}{37}.

 

Let's call the center of the circle that segment AB is the diameter of O. Note that \triangle ODE is an isosceles right triangle. Solving for side OE, using the Pythagorean theorem, we find it to be 5\sqrt 2. Calling the point where segment OE intersects circle O, the point I, segment IE would be 5\sqrt 2-2. Also, noting that \triangle ADE is a right triangle, we solve for side AE, using the Pythagorean Theorem, and get \sqrt {74}. Using Power of Point on point E, we can solve for CE. We can subtract CE from AE to find AC and then solve for CB using Pythagorean theorem once more: AE \cdot CE=\text{Diameter of circle} O+IE \cdot IE

\Rightarrow\sqrt{74}\left( CE\right)=\left( 5\sqrt{2}+2\right)\left( 5\sqrt{2}-2\right)\Rightarrow CE=\dfrac{23\sqrt{74}}{37},

AC=AE-CE\rightarrow AC=\sqrt{74}-\dfrac{23\sqrt{74}}{37}\Rightarrow AC=\dfrac{14\sqrt{74}}{37}.

Now to solve for CB: AB^{2}-AC^{2}=CB^{2}\rightarrow4^{2}+\dfrac{14\sqrt{74}^{2}}{37}=CB^{2}\Rightarrow CB=\dfrac{10\sqrt{74}}{37},

Note that \triangle ABC is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases AC and BC, we get the area of triangle ABC to be \frac {140}{37}.

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